PHY1002 · Week 5 · Electric Potential: Potential Difference and Electric Potential, Potential Differences in a Uniform Electric Field

Potential difference and potential in a uniform field

Electric potential, potential difference, the volt and the electron volt; ΔV = −E·d in a uniform field and problems on accelerating charges.

electric potentialpotential differenceelectric potential energyvoltelectron voltequipotential surfaceconservative fielduniform electric field

The electric field is a vector, and adding it component by component in every problem is tiring. The electric potential is a scalar: it has no direction, only a sign. This week you define it and learn its simplest form, the potential in a uniform field. The exam has two question types on this topic: “what is $V_B - V_A$ ?” and “a charge is accelerated from rest, what is its speed?”. Both take a few lines. What costs points is signs.

Golden rules:

$\Delta V = V_B - V_A = -\int_A^B \vec E\cdot d\vec s$ . The minus sign is part of the definition.
If you move along the field, the potential drops. Field lines run from high potential to low potential.
$\Delta U = q\,\Delta V$ . Always put in the sign of the charge.
Energy is conserved: $\Delta K + \Delta U = 0$ , so $\Delta K = -q\,\Delta V$ .
Motion perpendicular to the field does not change the potential. Only the component parallel to the field counts.

Potential energy and potential

When you move a test charge $q_0$ by a small $d\vec s$ in an electric field, the work done by the field is $\vec F\cdot d\vec s = q_0\vec E\cdot d\vec s$ . Because the electrostatic force is conservative, this work does not depend on the path, and a potential energy (electric potential energy) corresponds to it:

$\Delta U = -W_{\text{field}} = -q_0\int_A^B \vec E\cdot d\vec s$

This energy also depends on the test charge. We want a quantity independent of the test charge, so we define the potential energy per unit charge:

$V = \frac{U}{q_0}, \qquad \Delta V = \frac{\Delta U}{q_0} = -\int_A^B \vec E\cdot d\vec s$

Unit: $1\ \text{V} = 1\ \text{J/C}$ . From this the unit of the field follows too: $1\ \text{N/C} = 1\ \text{V/m}$ . On the exam the field is often given in V/m; they are the same thing.

Physical meaning: $\Delta V$ is the work an external agent must do to carry a $+1$ C charge from A to B at constant speed. The work of the external agent is $W_{\text{ext}} = q\,\Delta V$ , and the work of the field is $W_{\text{field}} = -q\,\Delta V$ . Pay attention to which one is asked.

PITFALL · Potential and potential energy are not the same thing. $V$ is a property of the place (in volts), $U$ is the energy of the charge at that place (in joules). At the same point a proton and an electron have the same potential, but their potential energies have opposite signs.

Only differences are physical. You choose the zero point: the ground in circuits, infinity for point charges.

The electron volt

On the atomic scale the joule is too large. The energy an electron (or proton) gains when it passes through a potential difference of 1 V is called an electron volt:

$1\ \text{eV} = (1.60\times10^{-19}\ \text{C})(1\ \text{V}) = 1.60\times10^{-19}\ \text{J}$

The convenience is this: if you multiply the charge in units of $e$ by the potential in volts, the energy comes out directly in eV. An alpha particle with charge $2e$ passing through 5.0 kV gains 10 keV.

Potential difference in a uniform field

If the field is the same everywhere (as between parallel plates), $\vec E$ comes out of the integral:

$\Delta V = -\vec E\cdot\int_A^B d\vec s = -\vec E\cdot\Delta\vec r = -(E_x\,\Delta x + E_y\,\Delta y + E_z\,\Delta z)$

If the displacement is parallel to the field and has length $d$ :

$\Delta V = -E\,d \qquad\text{(along the field)}, \qquad |\Delta V| = E\,d$

This relation also works the other way: given the voltage between plates and the distance, $E = |\Delta V|/d$ .

Equipotential surfaces

A surface made of points with the same potential is called an equipotential surface. If you walk perpendicular to the field, $\vec E\cdot d\vec s = 0$ and the potential does not change. The consequences:

Equipotential surfaces are perpendicular to the field lines everywhere.
In a uniform field the equipotentials are parallel planes, equally spaced.
No work is needed to move a charge along an equipotential surface.

TIP · Direction check. After you find the result, ask: “Did I go along the field?” If yes, $\Delta V$ must be negative. This one question catches most sign errors.

Motion of a charge and conservation of energy

If a charge moves under the electric force only, the total energy is conserved:

$K_i + qV_i = K_f + qV_f \quad\Longleftrightarrow\quad \Delta K = -q\,\Delta V$

For a charge starting from rest:

$v = \sqrt{\frac{2\,|q|\,|\Delta V|}{m}}$

The meaning of the sign: a released charge lowers its potential energy. A positive charge goes to low potential (along the field), a negative charge goes to high potential (against the field). Both speed up.

Charge	Direction it goes on its own	$\Delta V$	$\Delta U$	$\Delta K$
Positive	Along the field, low V	$< 0$	$< 0$	$> 0$
Negative	Against the field, high V	$> 0$	$< 0$	$> 0$

Worked examples

Example 1. Potential difference along the field. $\vec E = 600\,\hat i$ N/C. Point A is at the origin, point B at $(0.25, 0)$ m. What is $V_B - V_A$ ?

Δr = (0.25, 0) m
ΔV = -(Ex Δx + Ey Δy) = -(600)(0.25) - 0
ΔV = -150 V        (we went along the field, V dropped: consistent)

Example 2. Diagonal path. In the same field, point C is at $(0.25, 0.40)$ m. What is $V_C - V_A$ ?

Δr = (0.25, 0.40) m,  |Δr| = 0.47 m  (do NOT use this)
ΔV = -(600)(0.25) - (0)(0.40) = -150 V

The result is the same as in Example 1. The walk from B to C is perpendicular to the field, so B and C are on the same equipotential surface. Multiplying $E$ by the total distance is a classic mistake.

Example 3. Electron gun. An electron starts from rest and passes through a potential difference of 250 V. What are its final speed and kinetic energy?

The electron is negative: it speeds up toward high potential, ΔV = +250 V
ΔU = qΔV = (-1.60e-19)(250) = -4.0e-17 J
ΔK = -ΔU = +4.0e-17 J  = 250 eV
v = sqrt(2K/m) = sqrt(2(4.0e-17)/(9.11e-31)) = 9.37e6 m/s

A proton passing through the same potential difference gains the same energy (250 eV), but its mass is about 1840 times larger, so its speed is $\sqrt{1840}\approx 43$ times smaller.

Example 4. Slowing down. If a proton enters between the plates against the field, it slows down. The potential difference needed to stop it equals its kinetic energy in eV: a proton with 800 eV just manages to climb an 800 V hill. If the hill is higher, it turns back. This “think in eV” method reduces hard problems to two lines.

Example 5. Parallel plates and an intermediate point. Two large plates are 4.0 cm apart. The positive plate is at 200 V, the negative plate at 0 V (ground). (a) What is the field? (b) What is the potential at a point 1.0 cm from the positive plate? (c) At what speed does an electron released from rest at the negative plate hit the positive plate?

(a) E = ΔV/d = 200/0.040 = 5.0e3 V/m   (from the positive plate toward the negative)
(b) You went 1.0 cm along the field: V = 200 - (5.0e3)(0.010) = 150 V
(c) The electron goes from 0 V up to 200 V: K = 200 eV = 3.2e-17 J
    v = sqrt(2(3.2e-17)/(9.11e-31)) = 8.38e6 m/s

Between the plates the potential changes linearly. So for intermediate points, the “distance from the plate times E” method always works. Be careful which plate you measure from and what that plate’s potential is.

Example 6. Nonuniform field (a short look). If the field varies, for example $E_x = 400x^2$ (V/m, $x$ in meters), the formula $-E d$ does not apply and the integral is required:

$V(0.50) - V(0) = -\int_0^{0.50} 400x^2\,dx = -\frac{400}{3}(0.50)^3 = -16.7\ \text{V}$

The uniform-field formula is a special case of this general definition. If the field is given by a formula on the exam, your first step is to set up the integral.

Exam strategy

In potential difference problems, follow this order:

Mark the starting and ending points. $\Delta V = V_{\text{final}} - V_{\text{initial}}$ , the order matters.
Split the displacement vector into components. $\Delta\vec r = (\Delta x, \Delta y)$ .
Multiply only the component parallel to the field, and put in the minus sign. $\Delta V = -\vec E\cdot\Delta\vec r$ .
Do a direction check. If you went along the field, the result must be negative.
For an energy question, write $\Delta U = q\Delta V$ , then $\Delta K = -\Delta U$ . Thinking of the charge in eV speeds up the calculation; if a speed is asked, convert to joules at the end.

In speed questions pick the mass correctly: electron $9.11\times10^{-31}$ kg, proton $1.67\times10^{-27}$ kg. An electron and a proton passing through the same potential difference gain the same energy, but their speeds differ.

Common mistakes

Wrong idea	The truth
$\Delta V = +E d$ (along the field)	$\Delta V = -E d$ , the potential drops along the field
On a diagonal path $\Delta V = -E\,\lvert\Delta\vec r\rvert$	Only the component parallel to the field: $-\vec E\cdot\Delta\vec r$
An electron also goes to low potential	A negative charge goes to high potential
Skipping the sign of the charge in $\Delta U = q\Delta V$	For the electron $q = -1.60\times10^{-19}$ C
Potential and potential energy are the same	$V$ is J/C, $U$ is J; $U = qV$
$v = 2qV/m$ for the speed	There is a square root: $v = \sqrt{2qV/m}$
The work of the field and of the external agent are the same	Opposite signs: $W_{\text{ext}} = q\Delta V = -W_{\text{field}}$
$E = V/d$ without converting cm to meters	$d$ must be in meters, the result is in V/m
An alpha particle gets 5 keV at 5 kV	The charge is $2e$ : 10 keV

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

A uniform electric field of magnitude 450 N/C points in the $+x$ direction. Point A is at $x = 0$ and point B is at $x = 0.20$ m. What is $V_B - V_A$ ?

−90 V
+90 V
−2250 V
−9.0 V
0 V

Show answer

A) −90 V

You move along the field, so the potential drops. $V_B - V_A = -E\,d = -(450)(0.20) = -90$ V.
+90 V is a sign error, and −2250 V comes from dividing E by d.

Question 2easy

A proton starts from rest and is accelerated through a potential difference of 1500 V. What is its final speed? ( $m_p = 1.67\times10^{-27}$ kg)

2.9 × 10¹¹ m/s
5.4 × 10⁵ m/s
3.8 × 10⁵ m/s
2.3 × 10⁷ m/s
7.6 × 10⁵ m/s

Show answer

B) 5.4 × 10⁵ m/s

Conservation of energy: $\tfrac12 m v^2 = e\,|\Delta V|$ . $v = \sqrt{2(1.60\times10^{-19})(1500)/(1.67\times10^{-27})} = 5.36\times10^{5}$ m/s. $2.9\times10^{11}$ is forgetting the square root; $2.3\times10^7$ is using the electron mass.

Question 3easy

An electron moves from point A to point B, where $V_B - V_A = +120$ V. What is the change in the electron's electric potential energy?

+1.92 × 10⁻¹⁷ J
−1.92 × 10⁻¹⁷ J
+120 J
−7.5 × 10²⁰ J
0 J

Show answer

B) −1.92 × 10⁻¹⁷ J

$\Delta U = q\,\Delta V = (-1.60\times10^{-19})(+120) = -1.92\times10^{-17}$ J.
When a negative charge moves to a higher potential, its potential energy decreases and its kinetic energy increases.

Question 4medium

A uniform field $\vec E = 300\,\hat{i}$ N/C exists in a region. Point A is at $(0, 0)$ and point B is at $(0.40, 0.30)$ m. Find $V_B - V_A$ .

−150 V
−120 V
−90 V
+120 V
−210 V

Show answer

B) −120 V

$\Delta V = -\vec E\cdot\Delta\vec r = -(E_x\Delta x + E_y\Delta y) = -(300)(0.40) - 0 = -120$ V.
Only the component parallel to the field counts. −150 V uses the whole 0.50 m distance, and −90 V uses $\Delta y$ .

Question 5easy

Two large parallel plates are 2.0 cm apart and the potential difference between them is 240 V. What is the magnitude of the uniform electric field between the plates (in V/m)?

Show answer

1.2 × 10⁴ V/m

In a uniform field, $E = |\Delta V|/d = 240/0.020 = 1.2\times10^{4}$ V/m.
If you forget to convert centimeters to meters you get 120 V/m.

Question 6easy

In a region with a uniform electric field, a proton and an electron are each released from rest. Which statement is correct?

Both move toward lower potential.
Both move toward higher potential.
The proton moves toward lower potential; the electron moves toward higher potential.
The proton moves toward higher potential; the electron moves toward lower potential.
Neither moves, because the field is uniform.

Show answer

C) The proton moves toward lower potential; the electron moves toward higher potential.

A released charge moves in the direction that lowers its potential energy: $\Delta U = q\Delta V < 0$ .
For the proton $q>0$ , so $\Delta V<0$ (along the field). For the electron $q<0$ , so $\Delta V>0$ (against the field).

Question 7medium

An electron moving at $3.0\times10^{6}$ m/s enters a region where a uniform field of 500 N/C points in the same direction as the electron's velocity. How far does the electron travel before it momentarily stops?

2.6 cm
5.1 cm
10 cm
0.51 mm
1.0 m

Show answer

B) 5.1 cm

The force on the electron is opposite to the field, that is, opposite to its velocity: it slows down. $\tfrac12 m v^2 = eEd \Rightarrow d = \dfrac{m v^2}{2eE} = \dfrac{(9.11\times10^{-31})(9.0\times10^{12})}{2(1.60\times10^{-19})(500)} = 0.0512$ m.
This means the electron climbs a potential difference of 25.6 V. 2.6 cm comes from applying the factor 1/2 twice.

Question 8medium

An alpha particle (charge $+2e$ ) is accelerated from rest through a potential difference of 5.0 kV. What kinetic energy does it gain?

5.0 keV
10 keV
2.5 keV
8.0 × 10⁻¹⁶ J
5.0 × 10³ J

Show answer

B) 10 keV

$K = |q|\,|\Delta V| = (2e)(5000\ \text{V}) = 10\,000$ eV = 10 keV.
In joules, $10^4 \times 1.60\times10^{-19} = 1.6\times10^{-15}$ J. 5.0 keV forgets that the charge is $2e$ .

Question 9easy

A $+4.0\ \mu\text{C}$ charge is moved slowly from one point to another point on the same equipotential surface. How much work does the external agent do?

Depends on the path length
Positive, equal to qEd
Negative, equal to −qEd
Zero
Cannot be determined without E

Show answer

D) Zero

On an equipotential surface $\Delta V = 0$ , so $W = q\Delta V = 0$ .
The electric field is conservative; the work depends not on the path but only on the potentials at the end points.

Question 10medium

A $+3.0\ \mu\text{C}$ charge is moved at constant speed a distance of 0.15 m directly against a uniform field of $2.0\times10^{3}$ N/C. How much work does the external agent do?

−9.0 × 10⁻⁴ J
9.0 × 10⁻⁴ J
9.0 × 10⁻¹ J
4.0 × 10⁻² J
0 J

Show answer

B) 9.0 × 10⁻⁴ J

Moving against the field raises the potential: $\Delta V = +E d = (2000)(0.15) = 300$ V.
At constant speed, the work of the external force equals the increase in potential energy: $W = q\Delta V = (3.0\times10^{-6})(300) = 9.0\times10^{-4}$ J.
The negative answer is the work done by the field; the question asks for the external agent.

Question 11medium

Three equipotential surfaces at 10 V, 20 V and 30 V are flat, parallel and spaced 5.0 cm apart. What are the magnitude and direction of the electric field?

200 V/m, from the 10 V surface toward the 30 V surface
200 V/m, from the 30 V surface toward the 10 V surface
2.0 V/m, from 30 V toward 10 V
600 V/m, from 30 V toward 10 V
400 V/m, from 10 V toward 30 V

Show answer

B) 200 V/m, from the 30 V surface toward the 10 V surface

$E = |\Delta V|/d = 10\ \text{V}/0.050\ \text{m} = 200$ V/m.
The field always points in the direction in which the potential decreases: from 30 V to 10 V.
2.0 V/m is not converting 5.0 cm to meters.

Question 12hard

A proton with kinetic energy 2.0 keV moves into a region where it must climb a potential rise of 1500 V. What is its speed after the climb? ( $m_p = 1.67\times10^{-27}$ kg)

Show answer

3.1 × 10⁵ m/s

Work in electron volts: $K_f = K_i - e\Delta V = 2000\ \text{eV} - 1500\ \text{eV} = 500$ eV.
Convert to joules: $K_f = 500 \times 1.60\times10^{-19} = 8.0\times10^{-17}$ J. $v = \sqrt{2K_f/m_p} = \sqrt{2(8.0\times10^{-17})/(1.67\times10^{-27})} = 3.10\times10^{5}$ m/s.

Verification: All numerical answers were computed with node (verify/PHY1002/05.mjs): ΔV = −∫E·ds by a numerical sum with 10⁵ steps along the path, speeds and energies from conservation of energy.

Review card

dU = -q E·ds ΔU = -W_field V = U/q (1 V = 1 J/C) ΔV = VB - VA = -∫ E·ds (from A to B) Uniform field: ΔV = -E·Δr = -(Ex Δx + Ey Δy); distance d along the field: ΔV = -E d E = |ΔV|/d unit: V/m = N/C Go along the field and V DROPS. E points from high V to low V. ΔU = q ΔV (with the sign of q!) ΔK = -ΔU = -q ΔV Acceleration from rest: v = sqrt(2|q||ΔV|/m) 1 eV = 1.60e-19 J; q=2e, 5 kV -> 10 keV Equipotential surface ⟂ field line; work on it = 0 Work of the external agent (constant speed) = +qΔV; work of the field = -qΔV A positive charge goes to low V on its own, a negative charge to high V A step perpendicular to the field does NOT change the potential